Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung co

Question

Q. Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males. A person is selected at random from these 50 people and tested for lung related problems. Based on the given information, answer the following questions:

(i) What is the probability that selected person is a female?

(ii) If a male person is selected, what is the probability that he will not be suffering from lung problems?

(iii)

(a) A person selected at random is detected with lung complications. Find the probability that selected person is a female.

(OR)

(iii)

(b) A person selected at random is not having lung problems, find the probability that the person is a male.

CBSECBSE Class XII Board 2026Subjective· 4mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Conditional Probability

Conditional Probability

Roll a die and ask "what is the chance of an even number?" — that is $3/6$ . But suppose someone tells you the result is greater than 3. Now you are no longer looking at all six faces, only at $\{4,5,6\}$ , and two of those (4 and 6) are even, so the probability becomes $2/3$ . That change — from the probability of $A$ to the probability of $A$ given that $B$ has already occurred — is conditional probability.

The Idea: Shrink the Sample Space

Conditioning on $B$ throws away every outcome where $B$ is false and treats $B$ as the new "whole world." You measure $A$ only against what is still possible.

Note

Think of filtering a table of data: unconditional probability uses every row; conditional probability keeps only the rows where the condition is true.

The Definition

For events $A$ and $B$ with $P(B) > 0$ ,

P(A \mid B) = \frac{P(A \cap B)}{P(B)}.

We divide by $P(B)$ to rescale so that $B$ itself has probability $1$ ; the surviving part of $A$ is the overlap $A \cap B$ . Checking the die: $P(A \cap B) = P(\{4,6\}) = \tfrac{2}{6}$ and $P(B) = \tfrac{3}{6}$ , so $P(A \mid B) = \dfrac{2/6}{3/6} = \dfrac{2}{3}$ , matching the intuition.

Tip

Rearranging gives the multiplication rule $P(A \cap B) = P(A \mid B)\,P(B)$ , which is usually the easier way to compute a joint probability when a problem says "given that."

Two Cautions

$P(A \mid B)$ and $P(B \mid A)$ are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ . …

Part (b)Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

Conditioning on $B$ throws away every outcome where $B$ is false and treats $B$ as the new "whole world." You measure $A$ only against what is still possible.

Note

Think of filtering a table of data: unconditional probability uses every row; conditional probability keeps only the rows where the condition is true.

The Definition

For events $A$ and $B$ with $P(B) > 0$ ,

P(A \mid B) = \frac{P(A \cap B)}{P(B)}.

Tip

Rearranging gives the multiplication rule $P(A \cap B) = P(A \mid B)\,P(B)$ , which is usually the easier way to compute a joint probability when a problem says "given that."

Two Cautions

$P(A \mid B)$ and $P(B \mid A)$ are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ . …

Start your 14-day free trial to unlock the full solution →

(i) $P(F)=\tfrac25=0.4$ ; (ii) $P(L^c\mid M)=0.83$ ; (iii)(a) $P(F\mid L)=\tfrac{8}{25}=0.32$ ; (iii)(b) $P(M\mid L^c)=\tfrac{249}{425}\approx0.586$ .

This is a conditional-probability problem: we set up the events, use the Law of Total Probability to get the overall chance of lung complications, and Bayes' theorem to "reverse" a conditioning.

Events and given data. Let $M$ = male, $F$ = female, $L$ = lung complications, $L^c$ = no complications. Of the 50 smokers, 30 are male and 20 female, so

P(M)=\frac{30}{50}=\frac35=0.6,\qquad P(F)=\frac{20}{50}=\frac25=0.4.

The study gives $P(L\mid M)=\frac{170}{1000}=0.17$ and $P(L\mid F)=\frac{120}{1000}=0.12$ , hence $P(L^c\mid M)=0.83$ and $P(L^c\mid F)=0.88$ .

Total probability of complications.

P(L)=P(L\mid M)P(M)+P(L\mid F)P(F)=0.17(0.6)+0.12(0.4)=0.102+0.048=0.15,

so $P(L^c)=1-0.15=0.85$ .

Part (a)

(i) The person is chosen from the 50 smokers, 20 of whom are female:

P(F)=\frac{20}{50}=\frac25=0.4.

(ii) We want $P(L^c\mid M)$ , the chance a chosen male has no complications:

P(L^c\mid M)=1-P(L\mid M)=1-0.17=0.83.

(iii)(a) We are told the person has complications and asked for the chance they are female — a reverse conditional handled by Bayes' theorem:

$P(F\mid L)=\frac{P(L\mid F)\,P(F)}{P(L)}=\frac{0.12\times0.4}{0.15}=\frac{0.048}{0.15}=\frac{48}{150}=\frac{8}{25}=0.32.$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 88 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\dfrac{2}{3}$
›Reveal solutionSolution
The assertion is true: given the outcome is odd, the probability it is a prime is $\frac{2}{3}$ . The reason states the correct conditional probability formula. Since the reason directly justifies the calculation in the assertion, both are true and the reason is the correct explanation.

Concept first — Conditional probability asks: If we already know that event $B$ has occurred, what is the probability that event $A$ also occurs? The sample space shrinks from all possible outcomes to just those in $B$ . The formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is the precise way to compute this reduced probability.

Here, the die is unbiased, so each face $\{1,2,3,4,5,6\}$ has probability $\frac{1}{6}$ . The assertion involves two events:

$A$ : the number is prime. On a die, the primes are $2,3,5$ .

$B$ : the number is odd. The odd numbers are $1,3,5$ .

The condition "given that the number is odd" means we restrict attention to $B = \{1,3,5\}$ . Among these three equally likely outcomes, the primes are $3$ and $5$ — that's two out of three. So the conditional probability is $\frac{2}{3}$ .

Now let's verify step by step using the formula in Reason (R).

Define the events precisely.

$A = \{2,3,5\}$ , $B = \{1,3,5\}$ .

The sample space $S = \{1,2,3,4,5,6\}$ .

Compute $P(B)$ .

$B$ has 3 outcomes, each with probability $\frac{1}{6}$ , so $P(B) = \frac{3}{6} = \frac{1}{2}$ .

Compute $P(A \cap B)$ .

$A \cap B$ = numbers that are both prime and odd = $\{3,5\}$ . That's 2 outcomes, so $P(A \cap B) = \frac{2}{6} = \frac{1}{3}$ .

Apply the formula from Reason (R).

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/3}{1/2} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}.$
This matches the assertion exactly. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Since $S\cap F=F$ , the conditional probability $P(S\mid F)=1$ .

By the definition of conditional probability (with $P(F)\neq0$ ),

$P(S\mid F)=\frac{P(S\cap F)}{P(F)}.$ …
Start your 14-day free trial to see the full solution →
CBSE 2026Set CX1 markMCQ
$3P(A)=P(B)=\dfrac{5}{13}$
(a) $\dfrac{20}{39}$
(b) $\dfrac{16}{39}$
(c) $\dfrac{11}{39}$
(d) $\dfrac{14}{39}$
›Reveal solutionSolution

Using $P(A\cap B)=P(A/B)P(B)$ and the addition rule gives $P(A\cup B)=\dfrac{14}{39}$ — option (d).

Given: $3P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$ .

So $P(B)=\dfrac{5}{13}$ and $P(A)=\dfrac{1}{3}\cdot\dfrac{5}{13}=\dfrac{5}{39}$ .

Intersection (multiplication rule):

$P(A\cap B)=P(A/B)\,P(B)=\dfrac{2}{5}\cdot\dfrac{5}{13}=\dfrac{2}{13}.$

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$P(A) = \frac{7}{13},\ P(B) = \frac{9}{13},\ P(A \cap B) = \frac{4}{13} \Rightarrow P(A/B) =$
(a) $\frac{4}{9}$
(b) $\frac{4}{7}$
(c) $\frac{12}{13}$
(d) $\frac{1}{6}$
›Reveal solutionSolution

$P(A|B) = \frac{4}{9}$ .

Use the conditional-probability definition:

$P(A|B) = \frac{P(A\cap B)}{P(B)}.$

Substitute the given values: …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$P(B)=0.5$
(a) $\frac{15}{23}$
(b) $\frac{16}{25}$
(c) $\frac{16}{27}$
(d) $\frac{16}{23}$
›Reveal solutionSolution

By the definition of conditional probability, $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{16}{25}$ .

The conditional probability of $A$ given $B$ is defined as

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B) eq0$

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?$
›Reveal solutionSolution

List the equally likely outcomes for two children, restrict to those with at least one boy, then find the fraction that are both boys.

Sample space $=\{BB, BG, GB, GG\}$ , each equally likely.

Given at least one boy: reduced sample space $=\{BB,BG,GB\}$ (3 outcomes).

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$P(A) = 0.8$
(a) 0.8
(b) 0.5
(c) 0.32
(d) 0.4
›Reveal solutionSolution

Use the multiplication rule of conditional probability: $P(A\cap B) = P(B\mid A)\cdot P(A)$ .

Given $P(A)=0.8$ , $P(B\mid A)=0.4$ .

$P(A\cap B) = P(B\mid A)\cdot P(A) = 0.4 \times 0.8 = 0.32$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$P(A) = \dfrac{3}{10}$
(a) $\dfrac{1}{4}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{5}{12}$
(d) $\dfrac{7}{12}$
›Reveal solutionSolution

Find $P(A\cap B)$ from the addition rule, then use the conditional probability formula.

Given $P(A)=\dfrac{3}{10}$ , $P(B)=\dfrac{2}{5}$ , $P(A\cup B)=\dfrac{3}{5}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$\dfrac{3}{10}, \dfrac{1}{5}, \dfrac{1}{10}$
›Reveal solutionSolution

The problem statement itself states that if the doctor comes by other means, he will never be late.

Let $E_1,E_2,E_3,E_4$ denote the events that the doctor comes by train, bus, scooter, or other means, with $P(E_1)=\dfrac{3}{10}$ , $P(E_2)=\dfrac{1}{5}$ , $P(E_3)=\dfrac{1}{10}$ , $P(E_4)=\dfrac{2}{5}$ (these sum to 1, confirming they form a complete set of cases).

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. If P(A) = 1/2, P(B) = 3/8 and P(A\cupB) = 27/40 then P(A/B) is equal to:$
(a) 2/5
(b) 8/15
(c) 2/3
(d) 5/8
›Reveal solutionSolution

First find $P(A\cap B)$ using the addition rule, then apply the conditional probability formula $P(A/B)=\frac{P(A\cap B)}{P(B)}$ .

Given $P(A) = \frac12$ , $P(B)=\frac38$ , $P(A\cup B) = \frac{27}{40}$ .

Using the addition rule $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ :

$P(A\cap B) = P(A)+P(B)-P(A\cup B) = \frac{20}{40}+\frac{15}{40}-\frac{27}{40} = \frac{8}{40} = \frac15$

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4 then P(A\capB) = 0.32 Reason (R): Conditional Probability of 'B' when event A has occurred is given by P(B/A) = P(A\capB) / P(A)$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Both Assertion (A) and Reason (R) are false.
›Reveal solutionSolution

Applying the conditional-probability formula in R directly to the given numbers reproduces the value claimed in A.

Reason (R): $P(B/A) = \dfrac{P(A\cap B)}{P(A)}$ is the standard definition of conditional probability — true.

Checking Assertion (A): Given $P(A)=0.8,\ P(B/A)=0.4$ . Rearranging R's formula:

$P(A\cap B) = P(B/A)\times P(A) = 0.4 \times 0.8 = 0.32.$ …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$P(A) = \dfrac{1}{2}$
(a) 0
(b) $\dfrac{1}{2}$
(c) 1
(d) Not defined
›Reveal solutionSolution

Conditional probability is defined as $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ , which requires dividing by $P(B)$ — impossible when $P(B)=0$ .

Given $P(A)=\dfrac12$ , $P(B)=0$ .

By definition:

$P(A\mid B) = \frac{P(A\cap B)}{P(B)}$

…

Start your 14-day free trial to see the full solution →

Part (a)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days